Let A,B and C be three events, which are pair-wise independent and `bar(E )` denotes the complement of an event E. If `P (A nn B nn C) = 0 and PC) gt 0`, then `P[(bar(A) nn bar(B)) |C]` is equal to :

Options

A. `P(A) + P (bar(B))`
B. `P(bar(A)) - P (bar(B))`
C. `P(bar(A)) P (B)`
D. `P(bar(A)) + P (bar(B))`

Answer

To solve the problem, we need to find \( P(\bar{A} \cap \bar{B} | C) \) given that events A, B, and C are pairwise independent and \( P(A \cap B \cap C) = 0 \) with \( P(C) > 0 \).

### Step-by-Step Solution:

1. **Understanding the Problem**: We need to find the conditional probability \( P(\bar{A} \cap \bar{B} | C) \). By the definition of conditional probability, we have:
\[ P(\bar{A} \cap \bar{B} | C) = \frac{P(\bar{A} \cap \bar{B} \cap C)}{P(C)} \]

**Hint**: Remember that conditional probability is defined as the probability of the intersection of events divided by the probability of the conditioning event.

2. **Using Independence**: Since A, B, and C are pairwise independent, we can express the probabilities of the complements:
\[ P(\bar{A} \cap \bar{B} \cap C) = P(\bar{A}) \cdot P(\bar{B}) \cdot P(C) \]

**Hint**: For independent events, the probability of the intersection is the product of their probabilities.

3. **Calculating Complements**: The probabilities of the complements can be expressed as:
\[ P(\bar{A}) = 1 - P(A) \quad \text{and} \quad P(\bar{B}) = 1 - P(B) \]

**Hint**: Remember that the probability of the complement of an event is equal to 1 minus the probability of the event itself.

4. **Substituting Back**: Now substitute these into the equation for \( P(\bar{A} \cap \bar{B} \cap C) \):
\[ P(\bar{A} \cap \bar{B} \cap C) = (1 - P(A))(1 - P(B))P(C) \]

**Hint**: Keep track of the terms while substituting to avoid mistakes.

5. **Final Expression for Conditional Probability**: Now substitute this back into the expression for conditional probability:
\[ P(\bar{A} \cap \bar{B} | C) = \frac{(1 - P(A))(1 - P(B))P(C)}{P(C)} \]

**Hint**: Notice that \( P(C) \) cancels out in the numerator and denominator.

6. **Simplifying the Expression**: After cancellation, we have:
\[ P(\bar{A} \cap \bar{B} | C) = (1 - P(A))(1 - P(B)) \]

**Hint**: This is a product of the probabilities of the complements of A and B.

### Conclusion:
Thus, the final answer is:
\[ P(\bar{A} \cap \bar{B} | C) = (1 - P(A))(1 - P(B)) \]